Solve the differential equation left[frac{e^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the differential equation: $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1$Rearranging the equation to t

Vedclass · Accepted Answer

$y e^{2 \sqrt{x}} = 2 \sqrt{x} + C$

Vedclass · Answer

(A) Given the differential equation: $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}ight] \frac{d x}{d y}=1$Rearranging the equation to the form $\frac{dy}{dx} + Py = Q$:$\frac{dx}{dy} = \frac{\sqrt{x}}{e^{-2 \sqrt{x}} - y}$$\frac{dy}{dx} = \frac{e^{-2 \sqrt{x}} - y}{\sqrt{x}} = \frac{e^{-2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}$Thus,$\frac{dy}{dx} + \frac{1}{\sqrt{x}} y = \frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{\sqrt{x}}$ and $Q = \frac{e^{-2 \sqrt{x}}}{\sqrt{x}}$.Calculating the Integrating Factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int \frac{1}{\sqrt{x}} dx} = e^{2 \sqrt{x}}$The general solution is given by $y(I.F.) = \int (Q 	imes I.F.) dx + C$:$y e^{2 \sqrt{x}} = \int \left( \frac{e^{-2 \sqrt{x}}}{\sqrt{x}} 	imes e^{2 \sqrt{x}} ight) dx + C$$y e^{2 \sqrt{x}} = \int \frac{1}{\sqrt{x}} dx + C$$y e^{2 \sqrt{x}} = 2 \sqrt{x} + C$

Solve the differential equation $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{dy}=1$ where $x \neq 0$.

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $(1+x^2) \frac{dy}{dx} + y = e^{\tan^{-1} x}$,with $y(1)=0$. Then $y(0)$ is

Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx}-\sin(2y)=x^{3}(2-x^{3})\cos^{2}y,$ for $x\ne0.$ If $y(2)=0,$ then $\tan(y(1))$ is equal to

Let the solution $y=y(x)$ of the differential equation $\frac{dy}{dx}-y=1+4 \sin x$ satisfy $y(\pi)=1$. Then $y\left(\frac{\pi}{2}\right)+10$ is equal to:

The solution of $(2y - x) \frac{dy}{dx} = 1$ is

If $y=y(x)$ is the solution curve of the differential equation $x^{2} dy + (y - \frac{1}{x}) dx = 0$ for $x > 0$ and $y(1) = 1$,then $y(\frac{1}{2})$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module